Nuclear Medicine Exam Questions & Answers 2026 (1–10)

Question 1

Which detector type is most commonly used in a gamma camera?

TheoryPractice · Accepted Answer

Sodium iodide thallium-activated [NaI(Tl)]. NaI(Tl) crystals are the standard detector in Anger gamma cameras because of their high light output at 140 keV (the primary energy used in nuclear medicine) and well-established manufacturing. Semiconductor detectors like CZT are used in dedicated cardiac cameras but not the primary gamma camera platform.

Question 2

A parallel-hole collimator is used for a bone scan. What is the primary advantage of this collimator type?

TheoryPractice · Accepted Answer

Maintains image size independent of source-to-collimator distance. Parallel-hole collimators produce an image that is the same size as the object regardless of distance, making them ideal for whole-body imaging. Converging collimators magnify, diverging collimators minify, and pinhole collimators magnify only small fields of view.

Question 3

A NaI(Tl) gamma camera has a measured photopeak FWHM of 21 keV at 140 keV. What is the percent energy resolution?

TheoryPractice · Accepted Answer

15.0%. Percent energy resolution = (FWHM / photopeak energy) × 100 = (21 / 140) × 100 = 15%. Typical NaI(Tl) cameras achieve 9–12% at 140 keV; 15% indicates some degradation but is still within broad acceptance limits.

Question 4

Intrinsic spatial resolution of a gamma camera is measured WITHOUT the collimator. Which factor most directly degrades intrinsic resolution?

TheoryPractice · Accepted Answer

Statistical spread of light photons in the crystal. Intrinsic resolution is limited primarily by the statistical spread of scintillation photons within the NaI crystal, which creates uncertainty in locating the interaction. PMT diameter and electronics also contribute, but the photon statistics in the crystal are the dominant factor for intrinsic resolution.

Question 5

For imaging high-energy photons (e.g., 364 keV from I-131), which collimator is required?

TheoryPractice · Accepted Answer

High-energy general purpose (HEGP). I-131 emits 364 keV photons, which require thick septa to prevent penetration. High-energy general purpose (HEGP) collimators have thick lead septa designed for photon energies above ~300 keV. LEHR is for Tc-99m (140 keV) only.

Question 6

What is the primary function of the photomultiplier tube (PMT) in a gamma camera?

TheoryPractice · Accepted Answer

Convert light photons from the scintillator into an amplified electrical signal. PMTs detect the faint light (scintillation photons) produced when gamma rays interact in the NaI(Tl) crystal and amplify that light via successive dynode stages into a measurable electrical pulse. Position determination uses the relative outputs of multiple PMTs (Anger logic).

Question 7

In the Anger position logic circuit, how is the X, Y position of a gamma-ray interaction determined?

TheoryPractice · Accepted Answer

By weighted sum of all PMT outputs proportional to their distance from the event. Anger logic uses a resistor matrix (or digital equivalent) to compute weighted centroid positions. The X and Y coordinates are calculated from the fraction of light detected by each PMT relative to the total (Z-pulse), giving the centroid of the scintillation event.

Question 8

What is the purpose of the pulse height analyzer (PHA) in a gamma camera?

TheoryPractice · Accepted Answer

Accept only pulses within a set energy window to reject scatter. The PHA discriminates between pulses based on their amplitude (proportional to photon energy). By setting a window around the photopeak (e.g., ±10% of 140 keV), scatter photons with lower energy are rejected, improving image contrast.

Question 9

A gamma camera's observed count rate is 80,000 cps when the true count rate is 100,000 cps. What is the system dead time (τ) assuming a paralyzable model is NOT applicable and using the non-paralyzable formula n = m/(1 − mτ)?

TheoryPractice · Accepted Answer

2.5 µs. Non-paralyzable: n = m / (1 − mτ), so τ = (1 − m/n) / m = (1 − 80000/100000) / 80000 = 0.20 / 80000 = 2.5 × 10⁻⁶ s = 2.5 µs.

Question 10

SPECT imaging acquires data by rotating the gamma camera detector around the patient. What is the primary benefit over planar scintigraphy?

TheoryPractice · Accepted Answer

Three-dimensional tomographic images with improved lesion contrast. SPECT reconstructs true 3-D cross-sectional images, dramatically improving contrast-to-noise for small or deep lesions compared to planar projection imaging. Attenuation artifacts still exist in SPECT (corrected separately); sensitivity is not increased by removing the collimator.

Nuclear Medicine Exam Questions & Answers 2026 (1–10)

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